Question

A buffer was prepared by adding 1.00 mol of formic acid and 1.00 mol of sodium formate to 1.00 L of distilled water. A 100.0 mL aliquot of 1.00 M HCl is then added. What is the pH of the resulting buffer solution? (Formic acid pKa=3.74)

Answer 1

Buffer solution:

Buffer solution is made by the addition of acid and base solution. The pH of this solution is remain unchanged by adding acid or base.   In general the pH of this solution is constant.

This is an example of buffer solution. Thus use Henderson–hasselbalch equation.

pH = pKa + log( formate / formic acid)

Here moles of formate = formic acid = 1.00 mole

Volume = 1.00 L

Thus molarity = number of moles / volume in L

molarity of formate = molarity of formic acid = 1.00 M

Given that;

Formic acid pKa=3.74

moles HCl added = molarity * volume in L

= 0.100 L x 1.00 M = 0.1 00 Moles HCl

And
HCOO- + H+ >> HCOOH

moles HCOOH = 1.0 +0.100 = 1.10
moes HCOO- = 1.0 - 0.100 = 0.900

Therefore;

pH = pKa + log( formate / formic acid)

= 3.74 + log 0.900 /1.100

= 3.74+ (- 0.0872)

= 3.6528

= 3.65